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zimovet [89]
2 years ago
14

Need the following questions answered please and thank you

Chemistry
1 answer:
Ket [755]2 years ago
6 0
Hope this might help u

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Which law states that the volume and absolute temperature of a fixed quantity of gas are directly proportional under constant pr
seropon [69]

Charles’ Law........

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3 years ago
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Why is it good practice to perform a mixed melting point determination at two different ratios of unknown to known materials in
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The melting point of a particular compound is fixed and it is an important identification of an unknown compound. The practice to determine the melting point of an unknown material In different ratio with a known material is important to get the exact melting point of the unknown material. In different ratio the melting point of the unknown material will be fixed as the melting point of a pure material doesn't depend on the ratio in which they are mixed with other material. To get the exact melting point it is always good to get the melting point twice in different ratio.  

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Compare photons of gamma and radio radiation. which has: the longer wavelength? the greater frequency? the greater energy?
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<span>Gamma radiation has a shorter wavelength, a higher frequency and higher energy than radio radiation. Wavelength is inverse to frequency and energy (i.e. higher wavelength means lower frequency and lower energy, and vice versa).</span>
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3 years ago
D
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Answer:

Be, Mg, Ca, Sr are in the same group

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these elements are in alkaline earth metals

mark in as a brainly plz

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2 years ago
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The dilution factor D for an unseeded mixture of wastewater is 0.05. The DO of the mixture is initially 8.0 mg/L, and after 5 da
enyata [817]

Answer:

(i) 5-day BOD of the waste is 120 mg/l.

(ii) The ultimate carbonaceous BOD (Lo) is 20 mg/l.

Explanation:

The dilution factor D is 0.05.

The initial DO is 8.0 mg/L and the DO after 5 days is 2.0 mg/L.

The BOD of the waste for an unseeded mixture is

BOD_5=(DO_5-DO_0)/D=(8-2)/0.05=6/0.05=120mg/l

The ultimate carbonaceous BOD (Lo) can be calculated as

L = L_o*10 ^{- k_1t} \\L_o=L/10^{- k_1t}=2/10^{- 0.2*5}=2/10^{- 1}=2*10=20mg/l

6 0
2 years ago
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